Problem 48
Question
Guitar String A guitar string is pulled at point \(P\) a distance of 3 \(\mathrm{cm}\) above its rest position. It is then released and vibrates in damped harmonic motion with a frequency of 165 cycles per second. After 2 \(\mathrm{s}\) , it is observed that the amplitude of the vibration at point \(P\) is 0.6 \(\mathrm{cm}\) . (a) Find the damping constant \(c\) . (b) Find the damping constant \(c\) . above its rest position as a function of time. Take \(t=0\) to be the instant that the string is released.
Step-by-Step Solution
Verified Answer
The damping function is y(t) = 3e^{-0.8047t}cos(330πt).
1Step 1: Set up the damped harmonic motion model
The displacement is modeled by \(y(t) = Ae^{-kt}\cos(2\pi f t)\) where:
- \(A = 3\) cm (initial amplitude)
- \(f = 165\) Hz (frequency)
- \(k\) = damping constant (to be determined)
2Step 2: Find the damping constant k
After \(t = 2\) seconds, the amplitude is 0.6 cm:
\(3e^{-2k} = 0.6 \implies e^{-2k} = 0.2 \implies -2k = \ln(0.2)\)
\(k = \frac{-\ln(0.2)}{2} = \frac{\ln 5}{2} \approx 0.8047\)
\(3e^{-2k} = 0.6 \implies e^{-2k} = 0.2 \implies -2k = \ln(0.2)\)
\(k = \frac{-\ln(0.2)}{2} = \frac{\ln 5}{2} \approx 0.8047\)
3Step 3: Write the complete equation
\(y(t) = 3e^{-0.8047t}\cos(330\pi t)\)
where \(2\pi f = 2\pi(165) = 330\pi\).
where \(2\pi f = 2\pi(165) = 330\pi\).
Key Concepts
Guitar StringsFrequency of VibrationDamping ConstantHarmonic Motion Equation
Guitar Strings
Guitar strings are a classic example of objects that exhibit harmonic motion. When you pluck a guitar string, it vibrates as it moves up and down around its resting position. Each string's vibration creates sound waves which produce the music we hear. The movement of these strings can be described using the concepts of harmonic motion and damping.
- When you pluck a string, it is displaced from its resting position.
- As it vibrates, it passes energy into the surrounding air as sound.
- This energy loss causes the amplitude of vibration to gradually decrease.
Frequency of Vibration
The frequency of vibration refers to how often a guitar string oscillates. It is a significant factor in determining the pitch of the sound emitted. In the case of the given exercise, the guitar string vibrates at a frequency of 165 cycles per second, also noted as 165 Hertz (Hz).
- Higher frequency results in a higher pitched sound.
- The frequency depends on the properties of the string, such as its length and tension.
Damping Constant
The damping constant (
c
) is a measure of how quickly a vibrating object, like our guitar string, loses its energy over time. This constant plays a crucial role in determining the rate at which the amplitude of the vibration decreases. In the exercise, we need to calculate this damping constant to understand how quickly the vibration fades.
- A higher damping constant implies quicker energy loss and faster decrease in amplitude.
- In our example, after 2 seconds, the amplitude decreases from 3 cm to 0.6 cm.
Harmonic Motion Equation
Harmonic motion equations describe the motion of oscillating objects, and are particularly useful when dealing with harmonic and damped harmonic motion. The equation relevant to our exercise is given by:\[ x(t) = Ae^{-ct} \cos(2 \pi ft)\]
- \( x(t) \) represents the displacement of the string at any time \( t \).
- \( A \) is the initial displacement of the guitar string, which is 3 cm.
- \( c \) is the damping constant which we are solving for.
- \( f \) is the frequency of oscillation, which is 165 Hz in our case.
Other exercises in this chapter
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